ON REPRESENTATIONS OF QUANTUM GROUPSUq(f(K, H))

To Prof. Yingbo Zhang on the occasion of her 60th birthday

Xin Tang a,1 and Yunge Xu b,2

a. Department of Mathematics & Computer Science, Fayetteville State University,

Fayetteville, NC 28301, U.S.A E-mail: xtang@uncfsu.edub. Faculty of Mathematics & Computer Science, Hubei University,

Wuhan 430062, P.R.China E-mail: xuy@hubu.edu.cn

Abstract: As further generalizations of Uq(sl2), an interesting class of alge-bras Uq(f(K, H)) was introduced and studied in [10]. In this paper, we furtherstudy these algebras by realizing them as Hyperbolic algebras. Such a real-ization yields a natural construction of interesting families of irreducible rep-resentations of Uq(f(K, H)) by using methods in non-commutative algebraicgeometry([16]). We also investigate the relationship between Uq(f(K, H)) andthe algebras Uq(f(K)) introduced in [11]. As an application, we prove that anyfinite dimensional weight representation of Uq(f(K, H)) is completely reducible.In addition, we study the Whittaker model for the center of Uq(f(K, H)). Weprove that any Whittaker representation is irreducible if and only if it admitsa central character. Thus, a complete classification of all irreducible Whittakerrepresentations of Uq(f(K, H)) is obtained.

Keywords: Hyperbolic algebras, Spectral theory, Whittaker model, Quantumgroups

MSC(2000): 17B10, 17B35, 17B37

0. Introduction

Several works have been focused on the study of generalizations (or deforma-tions) of the quantized enveloping algebra Uq(sl2) ([17], [9], [11], [10], and [19]).Especially, a general class of algebras Uq(f(K)) (similar to Uq(sl2)) was introducedin [11], and their finite dimensional representations were studied as well. And therepresentation theory of these algebras was further investigated in [19] from theperspectives of spectral theory and Whittaker model. In [10], as generalizations ofthe algebras Uq(f(K)), a new class of algebras Uq(f(K, H)) was introduced andstudied. Note the Drinfeld quantum double of the positive part of the quantizedenveloping algebra Uq(sl2) studied in [7] or equivalently the two-parameter quan-tum groups Ur,s(sl2) studied in [2] features as a special case of these algebras. Thecondition on the parameter Laurent polynomial f(K, H) ∈ C[K±1,H±1] for theexistence of a Hopf algebra structure on Uq(f(K, H)) was determined, and finitedimensional irreducible representations were explicitly constructed as quotients ofhighest weight representations in [10]. This class of algebras provides a large familyof quantum groups in the sense of Drinfeld ([4]).

It is not surprising that these algebras share a lot of similar properties withthe two-parameter quantum groups Ur,s(sl2). So it might be useful to get more

1Xin Tang is grateful to Dr. A. Rosenberg for encouragement.2Yunge Xu is partially supported by NSFC under grant 10501010.

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information on the representation theory of these algebras. In the first part of thispaper, we will take a different approach to study the representations (not necessarilyfinite dimensional) of Uq(f(K, H)) from the point of view of spectral theory asdeveloped in [16]. Namely, we will realize these algebras as Hyperbolic algebras,then apply the general results obtained in [16] on Hyperbolic algebras to constructinteresting families of irreducible representations of Uq(f(K, H)). This approachyields the highest weight, the lowest weight and weight irreducible representations.

Note that there is also a close relationship between the representation theoryof Uq(f(K)) and that of Uq(f(K, H)). We will explore this relationship followingthe idea in [7]. As an application, we obtain some nice results on the categoryof all weight representations of Uq(f(K, H)). In particular, we are able to showthat the category of all weight representations of Uq(f(K, H)) is equivalent to theproduct of the category of weight representations of Uq(f(K)) with C∗ as tensorcategories. We have to admit that our results are motivated by the results in [7] forthe Drinfeld double, and we follow their approach closely. Combined with resultsproved in [11], we are able to show that any finite dimensional weight representationof Uq(f(K, H)) is completely reducible.

Finally, we study the Whittaker model of the center for these algebras. Wewill prove that any Whittaker representation is irreducible if and only if it admitsa central character. This result yields a complete classification of all irreducibleWhittaker representations of Uq(f(K, H)).

Now let us mention a little bit about the organization of this paper. In Section1, we recall the definition and basic facts about Uq(f(K)) and Uq(f(K, H)) from[11] and [10]. In Section 2, we recall some basic facts about spectral theory andHyperbolic algebras from [16]. Then we realize Uq(f(K, H)) as Hyperbolic algebrasand construct interesting families of irreducible weight representations. In Section3, we study the relationship between Uq(f(K)) and Uq(f(K, H)) from the perspec-tive of representation theory. In Section 4, we give a transparent description on thecenter of Uq(f(K, H)) and construct its Whittaker model. We study the Whittakerrepresentations of U(f(K, H)), and a complete classification of irreducible Whit-taker representations will be obtained. For the simplicity of computations, we willalways assume that f(K, H) = Km−Hm

q−q−1 and q is not a root of unity.

1. The algebras Uq(f(K, H))

Let C be the field of complex numbers and 0 6= q ∈ C such that q2 6= 1. Thequantized enveloping algebra corresponding to the simple Lie algebra sl2 is theassociative C−algebra generated by K±1, E, F subject to the following relations:

KE = q2EK, KF = q−2FK, KK−1 = K−1K = 1,

EF − FE =K −K−1

q − q−1.

This algebra is denoted by the standard notation Uq(sl2). It is well-known thatUq(sl2) is a Hopf algebra with the following Hopf algebra structure:

∆(E) = E ⊗ 1 + K ⊗ E, ∆(F ) = F ⊗K−1 + 1⊗ F ;

ε(E) = 0 = ε(F ), ε(K) = 1 = ε(K−1);

s(E) = −K−1E, s(F ) = −FK, s(K) = K−1.

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As generalizations of Uq(sl2), a class of algebras Uq(f(K)) parameterized by Lau-rent polynomials f(K) ∈ C[K, K−1] was later on introduced in [11]. For the reader’sconvenience, we recall their definition here.

Definition 1.1. (See [11]) For any Laurent polynomial f(K) ∈ C[K, K−1], Uq(f(K))is the C−algebra generated by E, F, K±1 subject to the following relations

KE = q2EK, KF = q−2FK;

KK−1 = K−1K = 1;

EF − FE = f(K).

The ring theoretic properties and finite dimensional representations were studiedin detail in [11]. We state some of the results here without proof. First of all,for the Laurent polynomials f(K) = a(Km − K−m) where a ∈ C∗ and m ∈ N,the algebras Uq(f(K)) have a Hopf algebra structure. In particular, we have thefollowing proposition quoted from [11].

Proposition 1.1. (Prop 3.3 in [11]) Assume f(K) is a non-zero Laurent polyno-mial in C[K, K−1]. Then the non-commutative algebra Uq(f(K)) is a Hopf algebrasuch that K, K−1 are group-like elements, and E,F are skew primitive elements ifand only if f(K) = a(Km−K−m) with m = t− s and the following conditions aresatisfied.

∆(K) = K ⊗K, ∆(K−1) = K−1 ⊗K−1;

∆(E) = Es ⊗ E + E ⊗Kt, ∆(F ) = K−t ⊗ F + F ⊗K−s;

ε(K) = ε(K−1) = 1, ε(E) = ε(F ) = 0;

S(K) = K−1, S(K−1) = K;

S(E) = −K−sEK−t, S(F ) = −KtFKs.

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For the case f(K) = Km−K−m

q−q−1 for m ∈ N and q is not a root of unity, thefinite dimensional irreducible representations were proved to be highest weight andconstructed explicitly in [11]. Furthermore, any finite dimensional representationsare completely reducible as stated in the following theorem from [11].

Theorem 1.1. (Thm 4.17 in [11]) With the above assumption for f(K) and q, anyfinite dimensional representation V of Uq(f(K)) is completely reducible.

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Remark 1.1. The representation theory of Uq(f(K)) was studied further from thepoints of views of spectral theory and Whittaker model in [19], where more familiesof interesting irreducible representations were constructed.

As generalizations of the algebras Uq(f(K)), another general class of algebrasparameterized by Laurent polynomials f(K, H) ∈ C[K±1,H±1] was also introducedand studied in [10]. First, let us recall the definition of Uq(f(K, H)) here:

Definition 1.2. (See [10]) Let f(K, H) ∈ C[K±1,H±1] be a Laurent polynomial,Uq(f(K, H)) is defined to be the C−algebra generated by E, F, K±1,H±1 subject

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to the following relations

KE = q2EK, KF = q−2FK,

HE = q−2EH, HF = q2FH,

KK−1 = K−1K = 1 = HH−1 = H−1H, KH = HK,

EF − FE = f(K, H).

It is easy to see that the Drinfeld quantum double of the positive part of Uq(sl2)or the two-parameter quantum group Ur,s(sl2) features as a special case amongthese algebras. The condition on the parameter f(K, H) for the existence of aHopf algebra structure on Uq(f(K, H)) was determined, and finite dimensionalirreducible representations were constructed explicitly as quotients of highest weightrepresentations in [10]. In addition, a counter example was also constructed to showthat not all finite dimensional representations are completely reducible in [10]. Soit would be interesting to know what kind of finite dimensional representations arecompletely reducible. We will address this question in Section 3.

2. Hyperbolic algebras and their representations

In this section, we realize Uq(f(K, H)) as Hyperbolic algebras and apply themethods in spectral theory as developed in [16] to construct irreducible weightrepresentations of Uq(f(K, H)). For the reader’s convenience, we need to recall alittle bit background about spectral theory and Hyperbolic algebras from [16].

2.1. Preliminaries on spectral theory. Spectral theory of abelian categorieswas first started by Gabriel in [5]. He defined the injective spectrum of any noe-therian Grothendieck category. This spectrum consists of isomorphism classes ofindecomposable injective objects. If R is a commutative noetherian ring, thenthe spectrum of the category of all R−modules is isomorphic to the prime spec-trum Spec(R) of R. And one can reconstruct any noetherian commutative scheme(X, OX) using the spectrum of the category of quasi-coherent sheaves of moduleson X. The spectrum of any abelian category was later on defined by Rosenbergin [16]. This spectrum works for any abelian category. Via this spectrum, one canreconstruct any quasi-separated and quasi-compact commutative scheme (X, OX)via the spectrum of the category of quasi-coherent sheaves of modules on X.

Though spectral theory is more important for the purpose of non-commutativealgebraic geometry, it has nice applications to representation theory. Spectrum hasa natural analogue of the Zariski topology and its closed points are in a one to onecorrespondence with the irreducible objects of the category. To study irreduciblerepresentations, one can study the spectrum of the category of all representations,then single out closed points of the spectrum with respect to the associated topol-ogy. As an application of spectral theory to representation theory, points of thespectrum ( hence representations ) have been constructed for a large family of alge-bras called Hyperbolic algebras in [16]. It is a pure luck that a large family of ‘small’algebras such as the first Weyl algebra A1, U(sl2) and their quantized versions anddeformations are Hyperbolic algebras.

Now we are going to review some basic notions and facts about spectrum of anyabelian category. First we review the definition of the spectrum of any abeliancategory, then we explain its applications in representation theory. We refer thereader to [16] for more details.

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Let CX be an abelian category and M,N ∈ CX be any two objects; We say thatM N if and only if N is a sub-quotient of the direct sum of finite copies of M .It is easy to verify that is a pre-order. We say M ≈ N if and only if M N andN M . It is obvious that ≈ is an equivalence. Let Spec(X) be the family of allnonzero objects M ∈ CX such that for any non-zero sub-object N of M , N M .

Definition 2.1. (See [16]) The spectrum of any abelian category is defined to be:

Spec(X) = Spec(X)/ ≈ .

Note that Spec(X) has a natural analogue of Zariski topology. Its closed pointsare in a one to one correspondence with irreducible objects of CX . To understandthe application of spectral theory to representation theory, let us look at an exam-ple. Suppose R is an associative algebra and CX is the category of all R−modules.Then closed points of Spec(X) are in a one to one correspondence to irreducibleR−modules. So the question of constructing irreducible R−modules is turned intoa more natural question of studying closed points of the spectrum. The advan-tage is that we can study the spectrum via methods in non-commutative algebraicgeometry. In addition, the left prime spectrum Specl(R) of any ring R is also de-fined in [16]. When R is commutative, Specl(R) is nothing but the prime spectrumSpec(R). We have to mention that it is proven in [16] that Specl(R) is isomorphicto Spec(R−mod) as topological spaces. So we will not distinguish the left primespectrum Specl(R) of a ring R and the spectrum Spec(R−mod) of the categoryof all R−modules.

2.2. Hyperbolic algebra Rξ, θ and its spectrum. Hyperbolic algebras arestudied by Rosenberg in [16] and by Bavula under the name of Generalized Weylalgebras in [1]. Hyperbolic algebra structure is very convenient for the constructionof points of the spectrum. And a lot of interesting algebras such as the first Weylalgebra A1, U(sl2) and their quantized versions have a Hyperbolic algebra structure.Points of the spectrum of the category of modules over Hyperbolic algebras areconstructed in [16]. We review some basic facts about Hyperbolic algebras and twoimportant construction theorems from [16].

Let θ be an automorphism of a commutative algebra R; and let ξ be an elementof R.

Definition 2.2. The Hyperbolic algebra Rθ, ξ is defined to be the R−algebragenerated by x, y subject to the following relations:

xy = ξ, yx = θ−1(ξ)

andxa = θ(a)x, ya = θ−1(a)y

for any a ∈ R. And Rθ, ξ is called a Hyperbolic algebra over R.

Let CX = CRθ,ξ be the category of modules over Rθ, ξ. We denote bySpec(X) the spectrum of CX . As we know that Spec(X) is isomorphic to the leftprime spectrum Specl(R). Points of the left prime spectrum of Hyperbolic algebrasare studied in [16], and in particular we have the following construction theoremsfrom [16].

Theorem 2.1. ( Thm 3.2.2.in [16] )(1) Let P ∈ Spec(R), and the orbit of P under the action of the θ is infinite.

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(a) If θ−1(ξ) ∈ P , and ξ ∈ P , then the left ideal

P1,1 : = P + Rθ, ξx + Rθ, ξy

is a two-sided ideal from Specl(Rθ, ξ).(b) If θ−1(ξ) ∈ P , θi(ξ) /∈ P for 0 ≤ i ≤ n − 1, and θn(ξ) ∈ P , then the

left ideal

P1,n+1 : = Rθ, ξP + Rθ, ξx + Rθ, ξyn+1

belongs to Specl(Rθ, ξ).(c) If θi(ξ) /∈ P for i ≥ 0 and θ−1(ξ) ∈ P , then

P1,∞ : = Rθ, ξP + Rθ, ξx

belongs to Specl(Rθ, ξ).(d) If ξ ∈ P and θ−i(ξ) /∈ P for all i ≥ 1, then the left ideal

P∞,1 : = Rθ, ξP + Rθ, ξy

belongs to Specl(Rθ, ξ).(2) If the ideal P in (b), (c) or (d) is maximal, then the corresponding left

ideal of Specl(Rθ, ξ) is maximal.(3) Every left ideal Q ∈ Specl(Rθ, ξ) such that θν(ξ) ∈ Q for a ν ∈ Z is

equivalent to one left ideal as defined above uniquely from a prime idealP ∈ Spec(R). The latter means that if P and P ′ are two prime ideals of Rand (α, β) and (ν, µ) take values (1,∞), (∞, 1), (∞,∞) or (1, n), then Pα,β

is equivalent to P ′ν,µ if and only if α = ν, β = µ and P = P ′.

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Theorem 2.2. (Prop 3.2.3. in [16])

(1) Let P ∈ Spec(R) be a prime ideal of R such that θi(ξ) /∈ P for i ∈ Z andθi(P )− P 6= Ø for i 6= 0, then P∞,∞ = Rξ, θP ∈ Specl(Rξ, θ).

(2) Moreover, if P is a left ideal of Rθ, ξ such that P ∩ R = P , then P =P∞,∞. In particular, if P is a maximal ideal, then P∞,∞ is a maximal leftideal.

(3) If a prime ideal P ′ ⊂ R is such that P∞,∞ = P ′∞,∞, then P ′ = θn(P ) for

some integer n. Conversely, θn(P )∞,∞ = P∞,∞ for any n ∈ Z.

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2.3. Realize Uq(f(K, H)) as Hyperbolic algebras. Let R be the sub-algebra ofUq(f(K, H)) generated by EF, K±1, H±1, then R is a commutative algebra. Wedefine an algebra automorphism θ : R −→ R of R by setting

θ(EF ) = EF + f(θ(K), θ(H)),

θ(K±1) = q∓2K±1,

θ(H±1) = q±2H±1.

It is easy to see that θ extends to an algebra automorphism of R. Furthermore,we have the following lemma:

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Lemma 2.1. The following identities hold:

E(EF ) = θ(EF )E,

F (EF ) = θ−1(EF )F,

EK = θ(K)E,

FK = θ−1(K)F,

EH = θ(H)E,

FH = θ−1(H)F.

Proof: We only verify the first one and the rest of them can be checked similarly.

E(EF ) = E(FE + f(K, H))= (EF )E + Ef(K, H)= (EF )E + f(θ(K), θ(H))E= (EF + f(θ(K), θ(H)))E= θ(EF )E.

So we are done. 2

From Lemma 2.1, we have the following result:

Proposition 2.1. Uq(f(K, H)) = Rξ = EF, θ is a Hyperbolic algebra with Rand θ defined as above.

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It easy to see that we have the following corollary:

Corollary 2.1. (See also Prop. 2.5 in [10]) Uq(f(K, H)) is noetherian domain ofGK-dimension 4.

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2.4. Families of irreducible weight representations of Uq(f(K, H)). Nowwe can apply the above construction theorems to the case of Uq(f(K, H)), andconstruct families of irreducible weight representations of Uq(f(K, H)).

To proceed, we first state a lemma:

Lemma 2.2. Let Mα,β,γ = (ξ − α, K − β, H − γ) ⊂ R be a maximal ideal of R,then θn(Mα,β,γ) 6= Mα,β,γ for any n ≥ 1. In particular, Mα,β,γ has an infinite orbitunder the action of θ.

Proof: We have

θn(K − β) = (q−2nK − β)= q−2n(K − q2nβ).

Since q is not a root of unity, q2n 6= 1 for any n 6= 0. So we have θn(Mα,β,γ) 6= Mα,β,γ

for any n ≥ 1. 2

Now we construct all irreducible weight representations of Uq(f(K, H)) for f(K, H) =a(Km−Hm) with a 6= 0 ∈ C and m ≥ 1. Without loss of generality, we may assumethat f(K) = Km−Hm

q−q−1 . We need another lemma:

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Lemma 2.3. (1) For n ≥ 0, we have the following:

θn(EF ) = EF +1

q − q−1(q−2m(1− q−2nm)

1− q−2mKm − q2m(1− q2nm)

1− q2mHm).

(2) For n ≥ 1, we have the following:

θ−n(EF ) = EF − 1q − q−1

(1− q2nm

1− q2mKm − 1− q−2nm

1− q−2mHm).

Proof: For n ≥ 1, we have

θn(EF ) = EF +1

q − q−1((q−2m + · · ·+ q−2nm)Km

− (q2m + · · ·+ q2nm)Hm)

= EF +1

q − q−1(q−2m(1− q−2nm)

1− q−2mKm − q2m(1− q2nm)

1− q2mHm).

The second statement can be verified similarly. 2

Theorem 2.3. (1) If α = βm−γm

q−q−1 , (β/γ)m = q2mn for some n ≥ 0, thenθn(ξ) ∈ Mα,β,γ and θ−1(ξ) ∈ Mα,β,γ , thus Uq(f(K, H))/P1,n+1 is a finitedimensional irreducible representation of Uq(f(K, H)).

(2) α = βm−γm

q−q−1 and (β/γ)m 6= q2mn for all n ≥ 0, then Uq(f(K, H))/P1,∞ isan infinite dimensional irreducible representation of Uq(f(K, H)).

(3) If α = 0 and 0 6= 1q−q−1 ( 1−q2nm

1−q2m βm − 1−q−2nm

1−q−2m γm) for any n ≥ 1, thenUq(f(K, H))/P∞,1 is an infinite dimensional irreducible representation ofUq(f(K, H)).

Proof: Since θ−1(ξ) = ξ − Km−Hm

q−q−1 , thus θ−1(ξ) ∈ Mα,β,γ if and only if α =βm−γm

q−q−1 . Now by the proof of Lemma 2.3, we have

θn(ξ) = ξ +1

q − q−1((q−2m + · · ·+ q−2nm)Km

− (q2m + · · ·+ q2nm)Hm)

= ξ +1

q − q−1(q−2m(1− q−2nm)

1− q−2mKm − q2m(1− q2nm)

1− q2mHm).

Hence θn(ξ) ∈ Mα,β,γ if and only if

0 = α +1

q − q−1((q−2m + · · · ,+q−2nm)βm − (q2m + · · · ,+q2nm)γm)

= α +1

q − q−1(q−2m(1− q−2nm)

1− q−2mβm − q2m(1− q2nm)

1− q2mγm).

Hence when α = βm−γm

q−q−1 , (β/γ)m = q2mn for some n ≥ 0, we have

θn(ξ) ∈ Mα,β,γ , θ−1(ξ) ∈ Mα,β,γ .

Thus by Theorem 2.1, Uq(f(K, H))/P1,n+1 is a finite dimensional irreducible rep-resentation of Uq(f(K, H)). So we have already proved the first statement, the restof the statements can be similarly verified. 2

Remark 2.1. The representations we constructed in Theorem 2.3 exhaust all fi-nite dimensional irreducible weight representations, the highest weight irreduciblerepresentations and the lowest weight irreducible representations of Uq(f(K, H)).

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Remark 2.2. Finite dimensional irreducible weight representations have been con-structed in [10] as quotients of highest weight representations. And a counter ex-ample has also been constructed in [10] to indicate that not all finite dimensionalrepresentations are completely reducible.

Apply the second construction theorem, we have the following theorem:

Theorem 2.4. If α 6= − 1q−q−1 ( q−2m(1−q−2nm)

1−q−2m βm − q2m(1−q2nm)1−q2m γm) for any n ≥ 0

and α 6= 1q−q−1 ( 1−q2nm

1−q2m βm − 1−q−2nm

1−q−2m γm) for any n ≥ 1, then Uq(f(K, H))/P∞,∞is an infinite dimensional irreducible weight representation of Uq(f(K, H)).

Proof: The proof is very similar to that of Theorem 2.3, we will omit it here.2

Corollary 2.2. The representations constructed in Theorem 2.3 and Theorem 2.4exhaust all irreducible weight representations of Uq(f(K, H)).

Proof: It follows directly from Theorems 2.1, 2.2, 2.3 and 2.4. 2

3. The relationship between Uq(f(K)) and Uq(f(K, H))

We will denote by f(K, H) the polynomial Km−Hm

q−q−1 , and by f(K) the Lau-

rent polynomial Km−K−m

q−q−1 . We compare the algebras Uq(f(K)) and Uq(f(K, H)).As a result, we will prove that any finite dimensional weight representation ofUq(f(K, H)) is completely reducible.

First of all, it is easy to see that we have the following lemma generalizing thesituation in [7]:

Lemma 3.1. The map which sends E to E, F to F , K±1 to K±1 and H±1 to K∓1

extends to a unique surjective Hopf algebra homomorphism π : Uq(f(K, H)) −→Uq(f(K).

Proof: The proof follows from the fact that the kernel of π is generated byK − H−1, which is a Hopf ideal of Uq(f(K, H)). Note that both Uq(f(K)) andUq(f(K, H)) are Hopf algebras under the assumption on f(K) and f(K, H). So weare done. 2

Our main goal in this section is to describe those representations M such thatEndUq(f(K,H))(M) = C. Since KH is in the center and invertible, it acts on theserepresentations by a non-zero scalar. As in [7], for each z ∈ C∗, we define a C−algebra homomorphism πz : Uq(f(K, H)) −→ Uq(f(K)) as follows:

πz(E) = zm2 E, πz(F ) = F ;

πz(K) = z12 K, πz(H) = z

12 K−1.

It is easy to see that πz is an algebra epimorphism with the kernel of πz beinga two-sided ideal generated by KH − z. But they may not necessarily be a Hopfalgebra homomorphism unless z = 1. Similar as in [7], we have the following lemma:

Lemma 3.2. For any representation M of Uq(f(K, H)) such that EndUq(f(K,H))(M) =C, there exists a unique z ∈ C∗ such that M is the pullback of a representation ofUq(f(K)) through πz as defined above. In particular, any such irreducible represen-tation of Uq(f(K, H)) is the pullback of an irreducible representation of Uq(f(K))through the algebra homomorphism πz for some z ∈ C∗.

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We use the notation in [7]. Let M be a representation of Uq(f(K)), we denoteby Mz the representation of Uq(f(K, H)) obtained as the pullback of M via πz.Let εz be the one dimensional representation of Uq(f(K, H)) which is defined bymapping the generators of Uq(f(K, H)) as follows:

εz(E) = εz(F ) = 0;

εz(K) = z12 , εz(H) = z

12 .

Then we have the following similar lemma as in [7]:

Lemma 3.3. Let 0 6= z ∈ C, and M be a representation of Uq(f(K)). Then Mz∼=

εz ⊗M1∼= M1 ⊗ εz. In particular, if 0 6= z′ ∈ C and N is another representation

of Uq(f(K)), then we have

Mz ⊗Nz′∼= (M ⊗N)zz′ .

Proof: The proof is straightforward. 2

Let M be a representation of Uq(f(K, H)), we say M is a weight representationif H and K are acting on M semisimply. Let C be the category of all weightrepresentations of Uq(f(K)) and C be the category of all weight representations ofUq(f(K, H)). Let C∗ be the tensor category associated to the multiplicative groupC∗, then we have the following:

Theorem 3.1. The category C is equivalent to the direct product of the categoriesC and C∗ as tensor categories.

Proof: The proof is similar to the one in [7], we refer the reader to [7] for moredetails. 2

Corollary 3.1. Any finite dimensional weight representation of Uq(f(K, H)) iscompletely reducible.

Proof: This follows from the above theorem and the fact that any finite dimen-sional representation of Uq(f(K)) is completely reducible (which is proved in [11]).2

Corollary 3.2. The tensor product of any two finite dimensional weight represen-tations is completely reducible.

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Remark 3.1. After the first draft of this paper was written, we have been kindlyinformed by J. Hartwig that the complete reducibility of finite dimensional weightrepresentations is also proved in his preprint [6] in a more general setting of Am-biskew polynomial rings via a different approach.

Remark 3.2. It might be interesting to study the decomposition of the productof two finite dimensional irreducible weight representations.

Remark 3.3. When m = 1, the above results are obtained in [7] for the Drin-feld double of the positive part of Uq(sl2), and equivalently for the two-parameterquantum groups Ur,s(sl2) in [2].

11

4. The Whittaker model for the center Z(Uq(f(K, H)))

Let g be a finite dimensional complex semisimple Lie algebra and U(g) be itsuniversal enveloping algebra. The Whittaker model for the center of U(g) was firststudied by Kostant in [12]. The Whittaker model for the center Z(U(g)) is definedby a non-singular character of the nilpotent Lie subalgebra n+ of g. Using theWhittaker model, Kostant studied the structure of Whittaker modules of U(g) anda lot of important results about Whittaker modules were obtained in [12]. Later on,Kostant’s idea was further generalized by Lynch in [13] and by Macdowell in [14] tothe case of singular characters of n+ and similar results hold in these generalizations.

The obstacle of generalizing the Whittaker model to the quantized envelopingalgebra Uq(g) with g of higher ranks is that there is no non-singular characterexisting for the positive part (Uq(g))≥0 of Uq(g) because of the quantum Serrerelations. In order to overcome this difficulty, it was Sevostyanov who first realizedto use the topological version Uh(g) over C[[h]] of quantum groups. Using a familyof Coxeter realizations Usπ

h (g) of the quantum group Uh(g) indexed by the Coxeterelements sπ, he was able to prove Kostant’s results for Uh(g) in [18]. While inthe simplest case of g = sl2, the quantum Serre relations are trivial, thus a directapproach should still work and this possibility has been worked out recently in [15].

In addition, it is reasonable to wonder whether the Whittaker model exists formost of the deformations of Uq(sl2), since they are so close to Uq(sl2) and share alot of properties with Uq(sl2). In this section, we show that there is such a Whit-taker model for the center of Uq(f(K, H)), and will study the Whittaker modulesfor Uq(f(K, H)). We obtain parallel results as in [12] and [15]. In deed, the ap-proach used here is more or less the same as the ones in [12] and [15]. For thereader’s convenience, we will work out all the details here. And we will use theterm Whittaker modules instead of Whittaker representations for convenience.

4.1. The center Z(Uq(f(K, H))) of Uq(f(K, H)). In this subsection, we first givea complete description of the center of Uq(f(K, H)). The center Z(Uq(f(K, H)))was also studied in [10] as well. However, the description of Z(Uq(f(K, H))) isincomplete there.

As mentioned from the very beginning, we always assume f(K, H) = Km−Hm

q−q−1

and q is not a root of unity. We define a Casimir element of Uq(f(H,K)) by setting:

Ω = FE +q2mKm + Hm

(q2m − 1)(q − q−1).

Then we have the following:

Proposition 4.1.

Ω = FE +q2mKm + Hm

(q2m − 1)(q − q−1)

= EF +Km + q2mHm

(q2m − 1)(q − q−1).

12

Proof: Since EF = FE + Km−Hm

q−q−1 , we have

Ω = FE +q2mKm + Hm

(q2m − 1)(q − q−1)

= EF − Km −Hm

q − q−1+

q2mKm + Hm

(q2m − 1)(q − q−1)

= EF +Km + q2mHm

(q2m − 1)(q − q−1).

So we are done. 2

In addition, we have the following lemma:

Lemma 4.1. Ω is in the center of Uq(f(K, H)).

Proof: It suffices to show that ΩE = EΩ,ΩF = FΩ,ΩK = KΩ,ΩH = HΩ.We will only verify that ΩE = EΩ and the rest of them are similar.

ΩE = (FE +q2mKm + Hm

(q2m − 1)(q − q−1))E

= (EF − Km −Hm

q − q−1+

q2mKm + K−m

(q2m − 1)(q − q−1))E

= E(FE) +Km + q2mHm

(q − q−1)(q2m − 1)E

= E(FE +q2mKm + Hm

(q2m − 1)(q − q−1))

= EΩ.

So we are done with the proof. 2

In particular, we have the following description of the center Z(Uq(f(K))) ofUq(f(K, H)), which refines a result in [10].

Proposition 4.2. (See also [10]) Z(Uq(f(K, H))) is the subalgebra of Uq(f(K, H))generated by Ω, (KH)±1. In particular, Z(Uq(f(K, H))) is isomorphic to the local-ization (C[Ω,KH])(KH) of the polynomial ring in two variables Ω,KH.

Proof: By Lemma 3.1., we have Ω, (KH)±1 ∈ Z(Uq(f(K, H))). Thus the sub-algebra C[Ω, (KH)±1] generated by Ω, (KH)±1 is contained in Z(Uq(f(K, KH))).So it suffices to prove the other inclusion Z(Uq(f(K, H))) ⊆ C[Ω, (KH)±1]. Notethat Uq(f(K, H)) =

⊕n∈Z≥0

Uq(f(K, H))n where Uq(f(K, H))n is the C−spanof elements u ∈ Uq(f(K, H)) | Ku = q2nuK, Hu = q−2nuH. Suppose x ∈Z(Uq(f(K, H))), then xK = Kx, xH = Hx. Thus x ∈ Uq(f(K))0 which is gen-erated by EF,K±1,H±1. By the definition of Ω, we know that Uq(f(K))0 is alsogenerated by Ω,K±1,H±1. Hence x =

∑fi,j(Ω)KiHj where fi,j(Ω) are polyno-

mials in Ω. Therefore

xE =∑

fi,j(Ω)KiHjE =∑

fi,j(Ω)q2i−2jEKiHj = Ex,

which forces i = j. So x ∈ C[Ω, (KH)±1] as desired. So we have proved thatZ(Uq(f(K, H))) = C[Ω, (KH)±1]. 2

13

4.2. The Whittaker model for Z(Uq(f(K, H))). Now we construct the Whit-taker model for Z(Uq(f(K, H))) following the lines in [12] and [15].

We first fix some notations. We denote by Uq(E) the subalgebra of Uq(f(K, H))generated by E, by Uq(F,K±1,H±1) the subalgebra of Uq(f(K, H)) generated byF,K±1,H±1. A non-singular character of the algebra Uq(E) is defined as follows:

Definition 4.1. An algebra homomorphism η : Uq(E) −→ C is called a non-singular character of Uq(E) if η(E) 6= 0.

From now on, we fix such a non-singular character of Uq(E) and denote it byη. Following [12], we define the concepts of a Whittaker vector and a Whittakermodule corresponding to the fixed non-singular character η.

Definition 4.2. Let V be a Uq(f(K, H))−module, a vector 0 6= v ∈ V is called aWhittaker vector of type η if E acts on v through the non-singular character η, i.e.,Ev = η(E)v. If V = Uq(f(K, H))v, then we call V a Whittaker module of type ηand v is called a cyclic Whittaker vector of type η.

The following decomposition of Uq(f(K, H)) is obvious:

Proposition 4.3. Uq(f(K, H)) is isomorphic to Uq(F,K±1,H±1) ⊗C Uq(E) asvector spaces and Uq(f(K, H)) is a free module over the subalgebra Uq(E).

2

Let us denote the kernel of η : Uq(E) −→ C by Uq,η(E), and we have the followingdecompositions of Uq(E) and Uq(f(K, H)).

Proposition 4.4. We have Uq(E) = C⊕ Uq,η(E). In addition,

Uq(f(K, H)) ∼= Uq(F,K±1,H±1)⊕ Uq(f(K, H))Uq,η(E).

Proof: It is obvious that Uq(E) = C⊕ Uq,η(E). And we have

Uq(f(K, H)) = Uq(F,K±1,H±1)⊗ (C⊕ Uq,η(E)),

thusUq(f(K, H)) ∼= Uq(F,K±1,H±1)⊕ Uq(f(K, H))Uq,η(E).

So we are done. 2

Now we define a projection:

π : Uq(f(K, H)) −→ Uq(F,K±1,H±1)

from Uq(f(K, H)) onto Uq(F,K±1,H±1) by taking the Uq(F,K±1,H±1)−componentof any u ∈ Uq(f(K, H)). We denote the image π(u) of u ∈ Uq(f(K, H)) by uη forshort.

Lemma 4.2. If v ∈ Z(Uq(f(K, H))) and u ∈ Uq(f(K, H)), then we have uηvη =(uv)η.

Proof: Let v ∈ Z(Uq(f(K, H))), u ∈ Uq(f(K, H)), then we have

uv − uηvη = (u− uη)v + uη(v − vη)= v(u− uη) + uη(v − vη),

which is in Uq(f(K, H))Uq,η(E). Hence (uv)η = uηvη. 2

14

Proposition 4.5. The map

π : Z(Uq(f(K, H)) −→ Uq(F,K±1,H±1)

is an algebra isomorphism of Z(Uq(f(K, H))) onto its image W (F,K±1,H±1) inUq(F,K±1,H±1).

Proof: It follows from that Lemma 4.2. that π is a homomorphism of algebras.Note that π(Ω) = η(E)F + q2mKm+Hm

(q2m−1)(q−q−1) with η(E) 6= 0. Furthermore, π(KH) =KH. Since Z(Uq(f(K, H)) = C[Ω, (KH)±1], so π is an injection. Thus π isan algebra isomorphism from Z(Uq(f(K, H))) onto its image W (F,K±1,H±1) inUq(F,K±1,H±1). 2

Lemma 4.3. If uη = u, then we have

uηvη = (uv)η

for any v ∈ Uq(f(K, H)).

Proof: We have

uv − uηvη = (u− uη)v + uη(v − vη)= uη(v − vη),

which is in Uq(f(K, H))Uq,η(E). So we have

uηvη = (uv)η

for any v ∈ Uq(f(K, H)). 2

Let A be the subspace of Uq(f(K, H)) spanned by K±i,H±i where i ∈ Z≥0.Then A is a graded vector space with

A[n] = CK±n ⊕ CH±n

for n ≥ 1, andA[0] = C,

andA[n] = 0

for n ≤ −1.As in [12] and [15], we define a filtration of Uq(F,K±1,H±1) as follows:

Uq(F,K±1,H±1)[n] = ⊕im+|j−k|≤nmUq(F,K±1,H±1)i,j,k

with Uq(F,K±1,H±1)i,j,k being the vector space spanned by F iKjHk. SinceW (F,K±1,H±1) is a subalgebra of Uq(F,K±1,H±1), it inherits a filtration fromUq(F,K±1,H±1). We denote by

W (F,K±1,H±1)[p] = C[(KH)±1]− span1,Ωη, · · · , (Ωη)pfor q ≥ 0. It is easy to see that

W (F,K±1,H±1)[p] ⊂ W (F,K±1,H±1)[p+1],

andW (F,K±1,H±1) =

∑p≥0

W (F,K±1,H±1)[p].

We have to mention that W (F,K±1,H±1)[p] give a filtration of W (F,K±1,H±1)which is compatible with the filtration of Uq(F,K±1,H±1). In particular,

W (F,K±1,H±1)[p] = W (F,K±1,H±1) ∩ Uq(F,K±1,H±1)[p]

15

for p ≥ 0.From the definition of Ω, we have the following description of η(Ω), the image

of Ω under η:

Lemma 4.4.

η(Ω) = η(E)F +q2mKm + Hm

(q2m − 1)(q − q−1).

2

Furthermore, we have the same useful lemma about Ωη as in [15] following fromdirect computations:

Lemma 4.5. Let x = Ωη and ξ = η(E). Then for n ∈ Z≥0. We have

xn = ξnFn + sn,

where sn ∈ Uq(F,K±1,H±1)[n] and is a sum of terms of the forms F iKjHk ∈Uq(F,K±1,H±1)[n] with i < n. In particular, sn are independent of Fn, wheneverit is nonzero.

2

Now we have the following decomposition of Uq(F,K±1,H±1).

Theorem 4.1. Uq(F,K±1,H±1) is free (as a right module) over W (F,K±1,H±1).And the multiplication induces an isomorphism

Φ: A⊗W (F,K±1,H±1) −→ Uq(F,K±1,H±1)

as right W (F,K±1,H±1)−modules. In particular, we have the following⊕p+lm=nm

A[p] ⊗W (F,K±1,H±1)[l] ∼= Uq(F,K±1,H±1)[n].

Proof: First of all, the map A × W (F,K±1,H±1) −→ Uq(F,K±1,H±1) isbilinear. So by the universal property of the tensor product, there is a map fromA⊗W (F,K±1,H±1) into Uq(F,K±1,H±1) defined by the multiplication. It is easyto check this map is a homomorphism of right W (F,K±1,H±1)−modules and issurjective as well. Now we show that it is injective. Let u ∈

⊕p+lm=nm A[p] ⊗

W (F,K±1,H±1)[l] with Φ(u) = 0. Up to an automorphism of W (F,K±1,H±1),we can write u =

∑i ui(K, H) ⊗ F i where ui(K, H) are Laurent polynomials in

C[K±1,H±1]. Then Φ(u) =∑

ui(K, H))F i = 0, so ui(K, H) = 0 for all i. Thusu = 0. So we have proved that Φ is an isomorphism of vector spaces. In addition,by counting the degrees of both sides, we also have⊕

p+lm=nm

A[p] ⊗W (F,K±1,H±1)[l] ∼= Uq(F,K±1,H±1)[n].

Thus we have proved the theorem. 2

Let Yη be the left Uq(f(K, H))−module defined by

Yη = Uq(f(K, H))⊗Uq(E) Cη,

where Cη is the one dimensional Uq(E)−module defined by the character η. It iseasy to see that

Yη = Uq(f(K, H))/Uq(f(K, H))Uq,η(E)

16

is a Whittaker module with a cyclic vector 1η = 1 ⊗ 1. Now we have a quotientmap from Uq(f(K, H)) to Yη as follows:

Uq(f(K, H)) −→ Yη is defined by u 7→ u1η,

for any u ∈ Uq(f(K, H)).If u ∈ Uq(f(K, H)), then there is a uη which is the unique element in Uq(F,K±1,H±1)

such that u1η = uη1η. As in [12], we define the η−reduced action of Uq(E) onUq(F,K±1,H±1) as follows:

x • v = (xv)η − η(x)v,

where x ∈ Uq(E) and v ∈ Uq(F,K±1,H±1). 2

Lemma 4.6. Let u ∈ Uq(f(K, H)) and x ∈ Uq(E), we have

x • uη = [x, u]η.

Proof: [x, u]1η = (xu− ux)1η = (xu− η(x)u)1η. Hence

[x, u]η = (xu)η − η(x)uη = (xuη)η − η(x)uη = x • uη.

2

Lemma 4.7. Let x ∈ Uq(E), u ∈ Uq(F,K±1,H±1), and v ∈ W (E,K±1,H±1),then we have

x • (uv) = (x • u)v.

Proof: Let v = wη for some w ∈ Z(Uq(f(K, H)), then uv = uwη = uηwη =(uw)η. Thus

x • (uv) = x • (uw)η = [x, uw]η

= ([x, u]w)η = [x, u]ηwη

= (x • uη)v= (x • u)v.

So we are done. 2

Let V be an Uq(f(K, H))−module and let Uq,V (f(K, H)) be the annihilator of Vin Uq(f(K, H)). Then Uq,V (f(K, H)) defines a central ideal ZV ⊂ Z(Uq(f(K, H)))by setting ZV = Uq,V (f(K, H)) ∩ Z(Uq(f(K, H))). Suppose that V is a Whit-taker module with a cyclic Whittaker vector w, we denote by Uq,w(f(K, H)) theannihilator of w in Uq(f(K, H)). It is obvious that

Uq(f(K, H))Uq,η(E) + Uq(f(K, H))ZV ⊂ Uq,w(f(K, H)).

In the next theorem, we show that the reverse inclusion holds. First of all, weneed an auxiliary Lemma:

Lemma 4.8. Let X = v ∈ Uq(F,K±1,H±1) | (x • v)w = 0, x ∈ Uq(E). Then

X = A⊗WV (F,K±1,H±1) + W (F,K±1,H±1),

where WV (F,K±1,H±1) = (ZV )η. In fact, Uq,V (F,K±1,H±1) ⊂ X and

Uq,w(F,K±1,H±1) = A⊗Ww(F,K±1,H±1),

where Uq,w(F,K±1,H±1) = Uq,w(f(K, H)) ∩ Uq(F,K±1,H±1).

17

Proof: Let us denote by Y = A⊗WV (F,K±1,H±1) + W (F,K±1,H±1) whereW (F,K±1,H±1) = (Z(Uq(f(K, H))))η. Thus we need to verify X = Y . Letv ∈ W (F,K±1,H±1), then v = uη for some u ∈ Z(Uq(f(K, H))). So we have

x • v = x • uη

= [x, u]η

= (xu)η − η(x)uη

= xηuη − η(x)uη

= 0.

So we have W (F,K±1,H±1) ⊂ X. Let u ∈ ZV and v ∈ Uq(F,K±1,H±1). Thenfor any x ∈ Uq(E), we have

x • (vuη) = (x • v)uη.

Since u ∈ ZV , then uη ∈ Uq,w(f(K, H)). Thus we have vuη ∈ X, hence

A⊗WV (F,K±1,H±1) ⊂ X,

which proves Y ⊂ X. Let A[i] be the subspace of C[K±1,H±1] spanned byK±i,H±i, and let WV (F,K±1,H±1) be the complement of WV (F,K±1,H±1) inW (F,K±1,H±1).

Let us setMi = A[i] ⊗WV (F,K±1,H±1),

thus we have the following:

Uq(F,K±1,H±1) = M ⊕ Y,

where M =∑

i≥1 Mi. We show that M ∩X 6= 0. Let M[k] =∑

1≤i≤k Mi, then M[k]

are a filtration of M . Suppose n is the smallest integer such that X ∩M[n] 6= 0 and0 6= y ∈ X∩M[n]. Then we have y =

∑1≤i≤n yi where yi ∈ Ai⊗WV (F,K±1,H±1).

Suppose we have chosen y in such a way that y has the fewest terms. By similarcomputations as in [15], we have 0 6= y − 1

ξ(q−2mn−1)E • y ∈ X ∩M[n] with fewerterms than y. This is a contradiction. So we have X ∩M = 0.

Now we prove that Uq,w(F,K±1,H±1) ⊂ X. Let u ∈ Uq,w(F,K±1,H±1) andx ∈ Uq(E), then we have xuw = 0 and uxw = η(x)uw = 0. Thus [x, u] ∈Uq,w(f(K, H)), hence [x, u]η ∈ Uq,w(F,K±1,H±1). Since u ∈ Uq,w(F,K±1,H±1) ⊂Uq,w(E,F,K±1,H±1), then x • u = [x, u]η. Thus x • u ∈ Uq,w(F,K±1,H±1). Sou ∈ X by the definition of X. Now we are going to prove the following:

W (F,K±1,H±1) ∩ Uq,w(F,K±1,H±1) = WV (F,K±1,H±1).

In fact, WV (F,K±1,H±1) = (ZηV ) and WV (F,K±1,H±1) ⊂ Uq,w(F,K±1,H±1).

So if v ∈ Ww(F,K±1,H±1) ∩ Uq,w(F,K±1,H±1), then we can uniquely writev = uη for u ∈ Z(Uq(f(K, H))). Then vw = 0 implies uw = 0 and hence u ∈Z(Uq(f(K, H)))∩Uq,w(F,K±1,H±1). Since V is generated cyclically by w, we haveproved the above statement. Obviously, we have Uq(f(K, H))ZV ⊂ Uq,w(f(K, H)).Thus we have A⊗WV (F,K±1,H±1) ⊂ Uq,w(F,K±1,H±1). Therefore, we have

Uq,w(F,K±1,H±1) = A⊗WV (F,K±1,H±1).

So we have finished the proof. 2

18

Theorem 4.2. Let V be a Whittaker module admitting a cyclic Whittaker vectorw, then we have

Uq,w(f(K, H)) = Uq(f(K, H))ZV + Uq(f(K, H))Uq,η(E).

Proof: It is obvious that

Uq(f(K, H))ZV + Uq(f(K, H))Uη(E) ⊂ Uq,w(f(K, H)).

Let u ∈ Uq,w(f(K, H)), we show that u ∈ Uq(f(K, H))ZV + Uq(f(K, H))Uq,η(E).Let v = uη, then it suffices to show that v ∈ A ⊗ WV (F,K±1,H±1). But v ∈Uq,w(F,K±1,H±1) = A⊗WV (F,K±1,H±1). So we have proved the theorem. 2

Theorem 4.3. Let V be any Whittaker module for Uq(f(K, H)), then the corre-spondence

V 7→ ZV

sets up a bijection between the set of all equivalence classes of Whittaker modulesand the set of all ideals of Z(Uq(f(K, H))).

Proof: Let Vi, i = 1, 2 be two Whittaker modules. If ZV1 = ZV2 , then clearly V1

is equivalent to V2 by the above Theorem. Now let Z∗ be an ideal of Z(Uq(f(K, H)))and let L = Uq(f(K, H))Z∗ + Uq(f(K, H))Uη(E). Then V = Uq(f(K, H))/L isa Whittaker module with a cyclic Whittaker vector w = 1. Obviously we haveUq,w(f(K, H)) = L. So L = Uq,w(f(K, H)) = Uq(f(K, H))ZV +Uq(f(K, H))Uq,η(E).This implies that

π(Z∗) = π(L) = π(ZV ).Since π is injective on Z(Uq(f(K, H))), thus ZV = Z∗. Thus we finished the proof.2

Theorem 4.4. Let V be an Uq(f(K, H))−module. Then V is a Whittaker moduleif and only if

V ∼= Uq(f(K, H))⊗Z(Uq(f(K,H)))⊗Uq(E) (Z(Uq(f(K, H)))/Z∗)η.

In particular, in such a case the ideal Z∗ is uniquely determined to be ZV .

Proof: If 1∗ is the image of 1 in Z(Uq(f(K, H)))/Z∗, then

AnnZ(Uq(f(K,H)))⊗Uq(F )(1∗) = Uq(E)Z∗ + Z(Uq(f(K, H)))Uq,η(E)

Thus the annihilator of w = 1⊗ 1∗ is

Uq,w(f(K, H)) = Uq(f(K, H))Z∗ + Uq(f(K, H))Uq,η(E)

Then the result follows from the last theorem. 2

Theorem 4.5. Let V be an Uq(f(K, H))−module with a cyclic Whittaker vectorw ∈ V . Then any v ∈ V is a Whittaker vector if and only if v = uw for someu ∈ Z(Uq(f(K, H))).

Proof: If v = uw for some u ∈ Z(Uq(f(K, H))), then it is obvious that v is aWhittaker vector. Conversely, let v = uw for some u ∈ Uq(f(K, H)) be a Whittakervector of V . Then v = uηw by the definition of Whittaker module. So we mayassume that u ∈ Uq(F,K±1,H±1). If x ∈ Uq(E), then we have xuw = η(x)uwand uxw = η(x)uw. Thus [x, u]w = 0 and hence [x, u]ηw = 0. But we havex • u = [x, u]η. Thus we have u ∈ X. We can now write u = u1 + u2 with u1 ∈Uq,w(F,K±1,H±1) and u2 ∈ W (F,K±1,H±1). Then u1w = 0. Hence u2w = v.

19

But u2 = uη3 with u3 ∈ Z(Uq(f(K, H))). So we have v = u3w which proves the

theorem. 2

Now let V be a Whittaker module and EndUq(f(K,H))(V ) be the endomorphismring of V as a Uq(f(K, H))−module. Then we can define the following homomor-phism of algebras using the action of Z(Uq(f(K, H))) on V :

πV : Z(Uq(f(K, H)) −→ EndUq(f(K,H))(V ).

It is clear that

Z(Uq(f(K, H)))/ZV (Uq(f(K, H))) ∼= πV (Z(Uq(f(K, H)))) ⊂ EndUq(f(K,H))(V ).

In fact, the next theorem says that this inclusion is an equality as well.

Theorem 4.6. Let V be a Whittaker Uq(f(K, H))−module. Then EndUq(f(K,H))(V ) ∼=Z(Uq(f(K, H)))/ZV . In particular, EndUq(f(K,H))(V ) is commutative.

Proof: Let w ∈ V be a cyclic Whittaker vector. If α ∈ EndUq(f(K,H))(V ),then α(w) = uw for some u ∈ Z(Uq(f(K, H))) by Theorem 4.5. Thus we haveα(vw) = vuw = uvw. Hence α = πu, which proves the theorem. 2

Now we are going to construct explicitly some Whittaker modules. Let

ξ : Z(Uq(f(K, H))) −→ C

be a central character of the center Z(Uq(f(K, H))). For any given central characterξ, let Zξ = Ker(ξ) ⊂ Z(Uq(f(K, H))) and Zξ is a maximal ideal of Z(Uq(f(K, H))).Since C is algebraically closed, then Zξ = (Ω−aξ,KH−bξ) for some aξ ∈ C, bξ ∈ C∗.For any given central character ξ, let Cξ,η be the one dimensional Z(Uq(f(K, H)))⊗Uq(E)−module defined by uvy = ξ(u)η(v)y for any u ∈ Z(Uq(f(K, H))) and anyv ∈ Uq(E). We set

Yξ,η = Uq(f(K, H))⊗Z(Uq(f(K,H)))⊗Uq(E) Cξ,η.

It is easy to see that Yξ,η is a Whittaker module of type η and admits a centralcharacter ξ. By Schur’s lemma, we know every irreducible representation has acentral character. As studied in [10], we know Uq(f(K, H)) has a similar theory forVerma modules. In fact, Verma modules also fall into the category of Whittakermodules if we take the trivial character of Uq(E). Namely we have the following

Mλ = Uq(f(K, H))⊗Uq(E,K±1,H±1) Cλ,

where K, H act on Cλ through the character λ of C[K±1,H±1] and Uq(E) acttrivially on Cλ. Thus, Mλ admits a central character. It is well-known that Vermamodules may not be necessarily irreducible, even though they have central charac-ters. However, Whittaker modules are in the other extreme as shown in the nexttheorem:

Theorem 4.7. Let V be a Whittaker module for Uq(f(K, H)). Then the followingstatements are equivalent.

(1) V is irreducible.

(2) V admits a central character.

(3) ZV is a maximal ideal.

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(4) The space of Whittaker vectors of V is one-dimensional.

(5) All nonzero Whittaker vectors of V are cyclic.

(6) The centralizer EndUq(f(K,H))(V ) is reduced to C.

(7) V is isomorphic to Yξ,η for some central character ξ.

Proof: It is easy to see that (2) − (7) are equivalent to each other by usingthe previous Theorems we have just proved. Since C is algebraically closed anduncountable, we also know (1) implies (2) by using a theorem due to Dixmier([3]). To complete the proof, it suffices to show that (2) implies (1), namely ifV has a central character, then V is irreducible. Let ω ∈ V be a cyclic Whit-taker vector, then V = Uq(f(K, H))ω. Since V has a central character, then itis easy to see from the description of the center that V has a C−basis consistingof elements Kiω, Hjω | i, j ∈ Z≥0. Let 0 6= v = (

∑ni=0 aiK

i +∑m

j=1 bjHj)ω ∈

V , then E(∑n

i=0 aiKi +

∑mj=1 bjH

j)ω = (∑n

i=0 q−2iaiKi +

∑mj=1 q2jbjH

j)Eω =ξ(

∑ni=0 q−2iaiK

i +∑m

j=1 q2jbjHj)ω. Thus we have 0 6= ξq−2nv−Ev ∈ V , in which

the top degree of K is n − 1. By repeating this operation finitely many times, wewill finally get an element 0 6= aω with a ∈ C∗. This means that V = Uq(f(K, H))vfor any 0 6= v ∈ V . So V is irreducible. Therefore, we are done with the proof. 2

It is easy to show the following two theorems, for more details about the proof,we refer the reader to [12].

Theorem 4.8. Let V be a Uq(f(K, H))−module which admits a central character.Assume that w ∈ V is a Whittaker vector. Then the submodule Uq(f(K, H))w ⊂ Vis irreducible.

2

Theorem 4.9. Let V1, V2 be any two irreducible Uq(f(K, H))−modules with thesame central character. If V1 and V2 contain Whittaker vectors, then these vectorsare unique up to scalars. And furthermore, V1 and V2 are isomorphic to each otheras Uq(f(K, H))−modules.

2

In fact we have the following description about the basis of an irreducible Whit-taker module (V,w) where w ∈ V is a cyclic Whittaker vector.

Theorem 4.10. Let (V,w) be an irreducible Whittaker module with a Whittakervector w, then V has a C−basis consisting of elements Kiω, Hjω | i, j ∈ Z≥0.

Proof: Since w is a cyclic Whittaker vector of the Whittaker module V , thenwe have V = Uq(f(K, H))w. Since Uq(f(K, H))w = Uq(F,K±,H±1) ⊗Uq(E) Cη,then we have V = Uq(F,K±1,H±1)w. Since V is irreducible, then V has a centralcharacter. Thus we have Ωw = λ(Ω)w. Now Ωw = (η(E)F + q2mKm+Hm

(q2m−1)(q−q−1) )w.Hence the action of F on V is uniquely determined by the action of K and H onV , and H−1v = aKv,K−1v = bHv for some a, b ∈ C∗ and for any v ∈ V . Thusthe theorem follows. 2

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